Strong Continuity

Strong continuity of a quantum channel family means continuity of each trace-class orbit in trace norm. It does not mean that one small time step is uniformly small for every state. An unbounded Hamiltonian already separates these notions, without dissipation or approximation. Naming both the space and the topology is therefore part of stating a dynamical theorem.

Required background. Trace Class Operators supplies the state Banach space. Helpful background. Quantum Dynamical Semigroups introduces the composition law.

Strong continuity concerns one trace-class orbit at a time

For bounded linear maps $\Phi_t$ on $\mathcal S_1(\mathcal H)$, strong continuity at zero means

$$\lim_{t\downarrow0}\|\Phi_t(X)-X\|_1=0 \quad\text{for every }X\in\mathcal S_1(\mathcal H).$$

Here “strong” is the strong operator topology for superoperators acting on the Banach space $\mathcal S_1$. It differs from the strong operator topology of observables acting on Hilbert vectors.

For a channel semigroup, contraction and the composition law propagate continuity from zero to every time. For $h\ge0$,

$$\|\Phi_{t+h}(X)-\Phi_t(X)\|_1 \le\|\Phi_h(X)-X\|_1.$$

The analogous estimate with $\Phi_{t-h}$ handles left continuity at $t>0$. Such a semigroup is called a $C_0$ semigroup.

Uniform, or operator-norm, continuity is the stronger requirement

$$\|\Phi_t-\mathrm{id}\|_{1\to1}\longrightarrow0.$$

Every norm-continuous semigroup has a bounded everywhere-defined generator, and conversely a bounded generator gives a norm-continuous exponential. A $C_0$ semigroup may have an unbounded generator defined only on a dense domain. In finite dimensions, these distinctions disappear because convergence on a finite basis controls the operator norm.

Hilbert-space continuity gives trace-norm continuity for unitary dynamics

Let $H$ be self-adjoint, possibly unbounded, and set $U_t=e^{-itH}$ with $\hbar=1$. The spectral theorem gives strong continuity of $U_t$ on $\mathcal H$. Define $\Phi_t(X)=U_tXU_t^\dagger$.

For a rank-one operator $|u\rangle\langle v|$,

$$\|\Phi_t(|u\rangle\langle v|)-|u\rangle\langle v|\|_1 \le\|U_tu-u\|\,\|v\| +\|u\|\,\|U_tv-v\|.$$

Thus the trace-norm orbit is continuous on finite-rank operators. Given arbitrary $X$, choose finite-rank $F$ with $\|X-F\|_1<\varepsilon$. Unitary invariance of trace norm yields

$$\|\Phi_t(X)-X\|_1 \le2\varepsilon+\|\Phi_t(F)-F\|_1.$$

Taking $t\to0$ and then $\varepsilon\to0$ proves the $C_0$ property. No finite energy or membership in $D(H)$ was required of the density operator. Those stronger conditions become relevant to differentiability, not to the existence and continuity of the orbit.

Worked example: arbitrarily fast phases prevent uniform continuity

Use an orthonormal basis $|k,0\rangle,|k,1\rangle$, $k\ge1$, and let

$$H|k,0\rangle=0,\qquad H|k,1\rangle=k|k,1\rangle.$$

This defines a self-adjoint operator on the domain whose excited coefficients satisfy $\sum_k k^2|b_k|^2<\infty$. For

$$|\psi_k\rangle=\frac{|k,0\rangle+|k,1\rangle}{\sqrt2}, \qquad t_k=\pi/k,$$

$U_{t_k}|\psi_k\rangle=(|k,0\rangle-|k,1\rangle)/\sqrt2$ is orthogonal to the input. Hence, with $\rho_k=|\psi_k\rangle\langle\psi_k|$,

$$\|\Phi_{t_k}(\rho_k)-\rho_k\|_1=2,\qquad \|\Phi_{t_k}-\mathrm{id}\|_{1\to1}=2.$$

Although $t_k\to0$, uniform continuity fails maximally along this sequence. Strong continuity still holds for each fixed $X$: the increasingly rapid phase occurs in a block that changes with $k$.

There is also a fixed bounded observable for which the dual orbit is not norm continuous. Let $A$ interchange $|k,0\rangle$ and $|k,1\rangle$ in every block. Then $\|A\|=1$, while the $k$th block gives $\|U_{t_k}^\dagger A U_{t_k}-A\|=2$. Thus continuity on trace-class states does not imply norm continuity of every orbit on $B(\mathcal H)$.

The normal dual evolves ultraweakly

For $T_t=\Phi_t^\dagger$, trace-class strong continuity implies

$$|\operatorname{Tr}[X(T_t(A)-A)]| \le\|\Phi_t(X)-X\|_1\,\|A\|\longrightarrow0.$$

This is point-ultraweak continuity of $T_t$. It controls every bounded-observable expectation in every normal initial state. It is enough for the normal dual description even when observable-norm continuity fails.

In the unitary example, $T_t(A)$ is also strongly continuous on Hilbert vectors, because both $U_t$ and $U_t^\dagger$ are strongly continuous and uniformly bounded. That additional statement still does not imply observable-norm continuity. The phrases “strongly continuous” and “continuous in norm” must therefore be attached to the specific action under discussion.

Exercises

A dense family of tests. Suppose channel maps satisfy $\Phi_t(F)\to F$ in trace norm for every finite-rank $F$. Prove strong continuity at zero on the whole trace class.

Solution

For fixed $X$, approximate it by finite-rank $F$. Channel contraction gives $\|\Phi_t(X)-X\|_1\le2\|X-F\|_1+\|\Phi_t(F)-F\|_1$. The order of choices matters: choose $F$ for the given input and tolerance, then choose the time.

Unbounded generator. In the block example compute the generator on $X_k=|k,0\rangle\langle k,1|$.

Solution

$\Phi_t(X_k)=e^{ikt}X_k$, so $\mathcal L(X_k)=ikX_k$. Since $\|X_k\|_1=1$ but $\|\mathcal L(X_k)\|_1=k$, the generator is unbounded. Each matrix unit is nevertheless a perfectly differentiable orbit, illustrating why a dense differentiability domain can coexist with an unbounded derivative.

References

• Engel, Klaus-Jochen, and Rainer Nagel. One-Parameter Semigroups for Linear Evolution Equations. Graduate Texts in Mathematics 194, Springer, 2000. doi:10.1007/b97696. Strong versus uniform continuity and the bounded-generator theorem.
• Hall, Brian C. Quantum Theory for Mathematicians. Graduate Texts in Mathematics 267, Springer, 2013. doi:10.1007/978-1-4614-7116-5. Self-adjoint generators and unitary evolution.